Power Series Expansion for Reciprocal of Cube Root of 1 + x

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Theorem

Let $x \in \R$ such that $-1 < x \le 1$.

Then:

$\dfrac 1 {\sqrt [3] {1 + x} } = 1 - \dfrac 1 3 x + \dfrac {1 \times 4} {3 \times 6} x^2 - \dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots$


Proof

\(\ds \frac 1 {\sqrt [3] {1 + x} }\) \(=\) \(\ds \paren {1 + x}^{-\frac 1 3}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-\frac 1 3}^{\underline k} } {k!} x^k\) General Binomial Theorem
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-\frac 1 3} - j} } {k!} x^k\) Definition of Falling Factorial
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {3 j - 1} } {3^k k!} x^k\) Translation of Index Variable of Product and simplifying
\(\ds \) \(=\) \(\ds 1 - \frac 1 3 x + \frac {1 \times 4} {3 \times 6} x^2 - \frac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots\) extracting the juice

$\blacksquare$


Sources